An acid `HA` ionizes as `HAhArrH^(+)+A^(-)` The `pH` of `1.0 M` solution is `5`. Its dissociation constant would be

Options

A. `1xx10^(-10)`
B. `5`
C. `5xx10^(-8)`
D. `1xx10^(-5)`

Explanation

To find the dissociation constant (Ka) of the acid HA, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions.
Given the pH of the solution is 5, we can calculate the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]
Substituting the given pH value:
\[ \text{[H⁺]} = 10^{-5} \, \text{M} \]

### Step 2: Determine the concentration of the acid (HA) and its dissociated ions.
The initial concentration of the acid HA is given as 1.0 M. When HA dissociates:
\[ HA \rightleftharpoons H^+ + A^- \]
Let \( \alpha \) be the degree of dissociation. At equilibrium:
- The concentration of H⁺ ions is [H⁺] = \( \alpha C \) = \( \alpha \times 1.0 \, \text{M} \)
- The concentration of A⁻ ions is also \( \alpha \times 1.0 \, \text{M} \)
- The concentration of undissociated HA is \( (1 - \alpha) \times 1.0 \, \text{M} \)

Since we know [H⁺] = \( 10^{-5} \, \text{M} \):
\[ \alpha \times 1.0 = 10^{-5} \]
Thus, \( \alpha = 10^{-5} \).

### Step 3: Calculate the concentration of A⁻ ions.
Since the dissociation produces equal amounts of H⁺ and A⁻:
\[ \text{[A⁻]} = \text{[H⁺]} = 10^{-5} \, \text{M} \]

### Step 4: Calculate the concentration of undissociated HA at equilibrium.
At equilibrium, the concentration of undissociated HA is:
\[ \text{[HA]} = 1.0 - 10^{-5} \approx 1.0 \, \text{M} \]
(We can approximate this because \( 10^{-5} \) is very small compared to 1.0.)

### Step 5: Use the expression for the dissociation constant (Ka).
The dissociation constant \( K_a \) is given by the formula:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Substituting the values we have:
\[ K_a = \frac{(10^{-5})(10^{-5})}{1.0} \]
\[ K_a = \frac{10^{-10}}{1.0} = 10^{-10} \]

### Final Answer:
The dissociation constant \( K_a \) of the acid HA is:
\[ K_a = 10^{-10} \]

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